Question

Find the \(L^{-1} (\frac{s}{2s+9+s^2})\).

(a) \(e^{-t} \{cos⁡(2\sqrt{2t})-sin⁡(\sqrt{2t})\}\)

(b) \(e^{-t} \{cos⁡(2\sqrt{2t})-sin⁡(2\sqrt{2t})\}\)

(c) \(e^{-t} \{cos⁡(2\sqrt{2t})-cos(\sqrt{2t})\}\)

(d) \(e^{-2t} \{cos⁡(2\sqrt{2t})-sin⁡(2\sqrt{2t})\}\)

The question was posed to me during an interview.

The origin of the question is General Properties of Inverse Laplace Transform in chapter Laplace Transform of Engineering Mathematics

Answer 1

Right answer is (b) \(e^{-t} \{cos⁡(2\sqrt{2t})-sin⁡(2\sqrt{2t})\}\)

The best I can explain: In the given question,

\(L^{-1} \left (\frac{s}{2s+9+s^2}\right )=L^{-1} \left (\frac{s}{(s+1)^2}+8)\right )\)

=\(L^{-1} \left (\frac{(s+1)-1}{(s+1)^2+8}\right )\)

=\(e^{-t} L^{-1} \left (\frac{(s-1)}{s^2+8}\right )\) ———————–By First Shifting Property

=\(e^{-t} L^{-1} \left (\frac{s}{s^2+8}\right )-e^{-t} L^{-1} \left (\frac{1}{s^2+8}\right )\)

=\(e^{-t} \{cos⁡(2\sqrt{2t})-sin⁡(2\sqrt{2t})\}\).